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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

My thoughts: $\displaystyle f(x)=\frac{x}{(x-1)(x+1)}$ is a function for which $f((-1,\ 1))=(-\infty,\ \infty)$.

- Thread starter Alexmahone
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- Thread starter
- #1

- Jan 26, 2012

- 268

My thoughts: $\displaystyle f(x)=\frac{x}{(x-1)(x+1)}$ is a function for which $f((-1,\ 1))=(-\infty,\ \infty)$.

- Jan 26, 2012

- 890

\(\displaystyle f(x)=\tan \left( \frac{x}{\pi/2} \right), \ \ x\in (-1,1) \)

My thoughts: $\displaystyle f(x)=\frac{x}{(x-1)(x+1)}$ is a function for which $f((-1,\ 1))=(-\infty,\ \infty)$.

will map \( (-1,1) \) to \( (-\infty, \infty) \) but \( [-1,1] \) might need a bit more thought, though you can just define \( f(-1)=f(1)=0 \)

CB

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